According to Newton’s law of cooling the rate of loss of heat, dQ/dt of the body is directly proportional to the different of temperature ΔT = (T_{2} – T_{1}) of the body and the surroundings. The law holds good only for small difference of temperature. Also the loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface.We can write

– dQ/dt = k (T_{2} – T_{1})

Where k is a positive constant depending upon the area and nature of the surface of the body. Suppose a body of mass m and specific s is at temperature T_{2 }Let T_{1} be the temperature falls by a small amount dT_{2} time dt, then the amount of heat lost is

dQ = msd T_{2}

**.·. **Rate of loss of heat is given by

dQ/ dt = ms = dT₂/dt

From equations we have

– ms dT₂/dt = k (T_{2} – T_{1})

dT₂ /T₂ – T₁ = = K/ms dt = – Kdt

Where K = k/ms

On integrating,

log_{e }(T_{2} – T_{1}) = – Kt + c

Or T_{2} = T_{1} + C’ e^{-kt }

Where C’ = e^{c}